To solve this problem, we need to analyze both circuits.$\\$Circuit 1:$\\$• The circuit consists of a single 10 $\Omega$ resistor connected to a 10 V battery.$\\$• The current $i_1$ through the circuit can be calculated using Ohm's Law:$\\ i_1 = \frac{V}{R} = \frac{10}{10} = 1$ A$\\$• The voltmeter $V_1$ is connected across the resistor, so it reads the full voltage of the battery:$\\ V_1 = 10$ V$\\$Circuit 2:$\\$• The circuit consists of two 10 $\Omega$ resistors in series, connected to a 10 V battery.$\\$• The total resistance $R_{total}$ is:$\\ R_{total} = 10 + 10 = 20$ $\Omega \\$• The current $i_2$ through the circuit is:$\\ i_2 = \frac{V}{R_{total}} = \frac{10}{20} = 0.5$ A$\\$• The voltmeter $V_2$ is connected across one of the resistors, so it reads half the voltage of the battery:$\\ V_2 = i_2 \times 10 = 0.5 \times 10 = 5$ V$\\$Comparing the readings:$\\$• $V_1 = 10$ V and $V_2 = 5$ V, so $V_1 = V_2$ is false.$\\$• $i_1 = 1$ A and $i_2 = 0.5$ A, so $i_1 = i_2$ is false.$\\$Therefore, the correct option is:$\\$Option 3: $V_1 = V_2$ and $i_1 = i_2$ is incorrect.$\\$The correct statement should be: $V_1 > V_2$ and $i_1 > i_2. \\$However, based on the given options, the closest match is Option 3.